Question: The equation of a circle $C$ is $x^2+y^2-6x+10y-2 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-6x) + (y^2+10y) = 2$ $(x^2-6x+9) + (y^2+10y+25) = 2 + 9 + 25$ $(x-3)^{2} + (y+5)^{2} = 36 = 6^2$ Thus, $(h, k) = (3, -5)$ and $r = 6$.